Question

a truck of mass 1000kg accelerated uniformly from rest to acquire velocity of 108 km/hr in 10 min . calculate the power developed by the engine of the truck

Answer 1

Answer:

Given:

Mass of truck = 1000 kg

Initial velocity = 0 m/s

Final Velocity = 108 km/hr

To find :

Power of engine of truck

Conversation:

We need to convert the velocity from km/hr to m/s.

108 km/hr = 108 × (5/18) = 30 m/s

Also time is to be converted from minutes to seconds;

10 minutes = 600 seconds.

Calculation:

Acceleration

= (30 - 0)/600

= 0.05 m/s².

Distance travelled

= ut + ½at²

= 0 + ½ × 0.05 × (600)²

= 9000 m

Force = m × a

=> Force = 1000 × 0.05 = 500 N

Therefore , Work = F × d

=> work = 500 × 9000

=> work = 45 × 10^5 joules

Power = work/time

=> P = (45 × 10^5)/600

=> P = 7.5 × 10³ watts

=> P = 7500 watts.

Answer 2

Question :-

→ Given above ↑

Answer :-

→ Power developed by truck is 7500 watt.

To Find :-

→ Power developed by engine of the truck .

Formula used :-

$\star \: \boxed{\sf{acceleration = \frac{v - u}{t} }} \\ \sf{ v = final \: velocuty \: } \\ \sf{u = initial \: velocity} \\ \\ \star \: \boxed{ \sf{ s = ut + \frac{1}{2} \: a \: {t}^{2} }} \\ \\ \star \: \boxed{ \sf{f = m \: a}} \\ \\ \star \: \boxed{ \sf{ w = \: f \: d} }\\ \\ \star \: \boxed{ \sf{ power = \frac{work}{time} }}$

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Explanation :-

Given that ,

• Mass of truck (m) = 1000 kg

• Initial velocity (u) = 0 , because it is at rest .
• final velocity (v) = 108 km/hr

• time (t) = 10 minutes or 600 second

Solution :-

Firstly we have to convert unit into m/s

$\because \\ \rightarrow \sf{1 \: \frac{km}{hr} = \frac{1000}{3600} \: \frac{m}{s} } \\ \therefore \: \\ \rightarrow \sf{ v \: = 108 \: \frac{km}{hr} = 108 \times \frac{5}{18} \: \: \frac{m}{s} } \\ \\ \rightarrow \sf{ v = 30 \: \: \frac{m}{s} } \:$

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Now calculate acceleration (a)

→ 10 minutes have 600 seconds ,

Apply the given formula of acceleration,

$\rightarrow \sf{a = \frac{30 - 0}{600} } \\ \\ \rightarrow \sf{ a \: = 0.05 \: \: \frac{m}{ {s}^{2} } }$

Now ,distance(s) travelled by truck is ,

$\rightarrow \sf{s = 0 + \frac{1}{2} \times 0.05 \times {(600)}^{2} } \\ \\ \rightarrow \sf{ s = 9000 \: m}$

truck travel 9000 m distance .

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We know that ,

→ Force (F) = m a

→ F = 1000 × 0.05

→ F = 50 N

Now ,

→ Work = Force(F) × distance (s)

→ Work = 50 × 9000

→ Work = 450 × 10 ⁴ Joule

Therefore ,

$\rightarrow \sf{power = \frac{work}{time} } \\ \\ \rightarrow \sf{power = \frac{450 \times {10}^{4} }{600} } \\ \\ \rightarrow \sf{power = 7.5 \times {10}^{3} } \\ \\ or \\ \rightarrow \boxed{\sf{power = 7500 \: watt}}$

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